Question

Assume that a,b,c and d are positive integers such that a^5=b⁴,c³=d²,and c-a=19 find d-b​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, a, b, c, d are positive integers such that

$\rm :\longmapsto\: {a}^{5} = {b}^{4}$

and

$\rm :\longmapsto\: {c}^{3} = {d}^{2}$

As

$\rm :\longmapsto\: {a}^{5} = {b}^{4}$

$\rm \implies\: {a}^{5} \: must \: be \: a \: perfect \: square$

Also,

$\rm :\longmapsto\: {c}^{3} = {d}^{2}$

$\rm \implies\: {c}^{3} \: must \: be \: a \: perfect \: square$

Again, it is given that

$\rm :\longmapsto\:c \: - \: a = 19$

Now, by hit and trial method, we have to choose values of c and a in such a way that they must be perfect squares and difference is 19.

So, By hit and trial

$\bf :\longmapsto\:c \: = \: 100 \: and \: a \: = \: 81$

Now,

$\rm :\longmapsto\: {a}^{5} = {b}^{4}$

On substituting the value of a, we get

$\rm :\longmapsto\: {81}^{5} = {b}^{4}$

$\rm :\longmapsto\: {( {3}^{4} )}^{5} = {b}^{4}$

$\rm :\longmapsto\: {3}^{20} = {b}^{4}$

$\bf\implies \:b = {3}^{5} \: \: \: \: \: as \: b > 0$

$\bf\implies \:b = 243$

Also,

$\rm :\longmapsto\: {c}^{3} = {d}^{2}$

On substituting the value of c, we get

$\rm :\longmapsto\: {100}^{3} = {d}^{2}$

$\rm :\longmapsto\: {( {10}^{2} )}^{3} = {d}^{2}$

$\rm :\longmapsto\: {10}^{6} = {d}^{2}$

$\bf\implies \:d = {10}^{3} \: \: \: \: \: as \: d > 0$

$\bf\implies \:d = 1000$

Now, Consider

$\rm :\longmapsto\:d - b$

$\rm \: = \: 1000 - 243$

$\rm \: = \: 757$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: d \: - \: b \: = \: 757 \: }}}$

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Additional Information :-

$\boxed{ \tt{ \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}$

$\boxed{ \tt{ \: {x}^{m} \div {x}^{n} = {x}^{m - n} \: }}$

$\boxed{ \tt{ \: {( {x}^{m} )}^{n} = {( {x}^{n} )}^{m} = {x}^{mn} \: }}$

$\boxed{ \tt{ \: {x}^{0} = 1 \: }}$

$\boxed{ \tt{ \: {x}^{ - n} = \frac{1}{ {x}^{n} } \: }}$